problem 31.7 (L)

Internal problem ID [13909]
Internal ﬁle name [OUTPUT/13081_Friday_February_23_2024_06_54_35_AM_38663670/index.tex]

Book: Ordinary Diﬀerential Equations. An introduction to the fundamentals. Kenneth B. Howell. second edition. CRC Press. FL, USA. 2020
Section: Chapter 31. Delta Functions. Additional Exercises. page 572
Problem number: 31.7 (L).
ODE order: 4.
ODE degree: 1.

\[ \boxed {y^{\prime \prime \prime \prime }-16 y=\delta \left (t \right )} \] With initial conditions \begin {align*} [y \left (0\right ) = 0, y^{\prime }\left (0\right ) = 0, y^{\prime \prime }\left (0\right ) = 0, y^{\prime \prime \prime }\left (0\right ) = 0] \end {align*}

Solving using the Laplace transform method. Let \[ \mathcal {L}\left (y\right ) =Y(s) \] Taking the Laplace transform of the ode and using the relations that \begin {align*} \mathcal {L}\left (y^{\prime }\right )&= s Y(s) - y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime }\right ) &= s^2 Y(s) - y'(0) - s y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime }\right ) &= s^3 Y(s) - y''(0) - s y'(0) - s^2 y \left (0\right )\\ \mathcal {L}\left (y^{\prime \prime \prime \prime }\right ) &= s^4 Y(s) - y'''(0) - s y''(0) - s^2 y'(0)- s^3 y \left (0\right ) \end {align*}

The given ode becomes an algebraic equation in the Laplace domain \[ s^{4} Y \left (s \right )-y^{\prime \prime \prime }\left (0\right )-s y^{\prime \prime }\left (0\right )-s^{2} y^{\prime }\left (0\right )-s^{3} y \left (0\right )-16 Y \left (s \right ) = 1\tag {1} \] But the initial conditions are \begin {align*} y \left (0\right )&=0\\ y^{\prime }\left (0\right )&=0\\ y^{\prime \prime }\left (0\right )&=0\\ y^{\prime \prime \prime }\left (0\right )&=0 \end {align*}

Substituting these initial conditions in above in Eq (1) gives \[ s^{4} Y \left (s \right )-16 Y \left (s \right ) = 1 \] Solving the above equation for \(Y(s)\) results in \[ Y(s) = \frac {1}{s^{4}-16} \] Applying partial fractions decomposition results in \[ Y(s)= -\frac {1}{32 \left (s +2\right )}+\frac {i}{32 s -64 i}-\frac {i}{32 \left (s +2 i\right )}+\frac {1}{32 s -64} \] The inverse Laplace of each term above is now found, which gives \begin {align*} \mathcal {L}^{-1}\left (-\frac {1}{32 \left (s +2\right )}\right ) &= -\frac {{\mathrm e}^{-2 t}}{32}\\ \mathcal {L}^{-1}\left (\frac {i}{32 s -64 i}\right ) &= \frac {i {\mathrm e}^{2 i t}}{32}\\ \mathcal {L}^{-1}\left (-\frac {i}{32 \left (s +2 i\right )}\right ) &= -\frac {i {\mathrm e}^{-2 i t}}{32}\\ \mathcal {L}^{-1}\left (\frac {1}{32 s -64}\right ) &= \frac {{\mathrm e}^{2 t}}{32} \end {align*}

Adding the above results and simplifying gives \[ y=-\frac {\sin \left (2 t \right )}{16}+\frac {\sinh \left (2 t \right )}{16} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\sin \left (2 t \right )}{16}+\frac {\sinh \left (2 t \right )}{16} \\ \end{align*}

Veriﬁcation of solutions

\[ y = -\frac {\sin \left (2 t \right )}{16}+\frac {\sinh \left (2 t \right )}{16} \] Warning, solution could not be veriﬁed

22.19.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime \prime \prime }-16 y=\mathit {Dirac}\left (t \right ), y \left (0\right )=0, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0, y^{\prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0, y^{\prime \prime \prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{t \hiderel {=}0\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=y^{\prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=y^{\prime \prime } \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=y^{\prime \prime \prime } \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} y_{4}^{\prime }\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & y_{4}^{\prime }\left (t \right )=\mathit {Dirac}\left (t \right )+16 y_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=y_{1}^{\prime }\left (t \right ), y_{3}\left (t \right )=y_{2}^{\prime }\left (t \right ), y_{4}\left (t \right )=y_{3}^{\prime }\left (t \right ), y_{4}^{\prime }\left (t \right )=\mathit {Dirac}\left (t \right )+16 y_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \\ y_{4}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 16 & 0 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right )+\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ \mathit {Dirac}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the forcing function}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{f}}\left (t \right )=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ \mathit {Dirac}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 16 & 0 & 0 & 0 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}^{\prime }\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \right )+{\moverset {\rightarrow }{f}} \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ], \left [2 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ -\frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-2, \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {1}{8} \\ \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{8} \\ \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \\ -\frac {1}{4} \\ \frac {\mathrm {I}}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{8} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ -\frac {\cos \left (2 t \right )}{4}+\frac {\mathrm {I} \sin \left (2 t \right )}{4} \\ \frac {\mathrm {I}}{2} \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{3}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (2 t \right )}{8} \\ -\frac {\cos \left (2 t \right )}{4} \\ \frac {\sin \left (2 t \right )}{2} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (2 t \right )}{8} \\ \frac {\sin \left (2 t \right )}{4} \\ \frac {\cos \left (2 t \right )}{2} \\ -\sin \left (2 t \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {General solution of the system of ODEs can be written in terms of the particular solution}\hspace {3pt} {\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right )+{\moverset {\rightarrow }{y}}_{p}\left (t \right ) \\ \square & {} & \textrm {Fundamental matrix}\hspace {3pt} \\ {} & \circ & \textrm {Let}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {be the matrix whose columns are the independent solutions of the homogeneous system.}\hspace {3pt} \\ {} & {} & \phi \left (t \right )=\left [\begin {array}{cccc} -\frac {{\mathrm e}^{-2 t}}{8} & \frac {{\mathrm e}^{2 t}}{8} & -\frac {\sin \left (2 t \right )}{8} & -\frac {\cos \left (2 t \right )}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{2 t}}{4} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{2 t}}{2} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} & \cos \left (2 t \right ) & -\sin \left (2 t \right ) \end {array}\right ] \\ {} & \circ & \textrm {The fundamental matrix,}\hspace {3pt} \Phi \left (t \right )\hspace {3pt}\textrm {is a normalized version of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {satisfying}\hspace {3pt} \Phi \left (0\right )=I \hspace {3pt}\textrm {where}\hspace {3pt} I \hspace {3pt}\textrm {is the identity matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\phi \left (t \right )\cdot \frac {1}{\phi \left (0\right )} \\ {} & \circ & \textrm {Substitute the value of}\hspace {3pt} \phi \left (t \right )\hspace {3pt}\textrm {and}\hspace {3pt} \phi \left (0\right ) \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} -\frac {{\mathrm e}^{-2 t}}{8} & \frac {{\mathrm e}^{2 t}}{8} & -\frac {\sin \left (2 t \right )}{8} & -\frac {\cos \left (2 t \right )}{8} \\ \frac {{\mathrm e}^{-2 t}}{4} & \frac {{\mathrm e}^{2 t}}{4} & -\frac {\cos \left (2 t \right )}{4} & \frac {\sin \left (2 t \right )}{4} \\ -\frac {{\mathrm e}^{-2 t}}{2} & \frac {{\mathrm e}^{2 t}}{2} & \frac {\sin \left (2 t \right )}{2} & \frac {\cos \left (2 t \right )}{2} \\ {\mathrm e}^{-2 t} & {\mathrm e}^{2 t} & \cos \left (2 t \right ) & -\sin \left (2 t \right ) \end {array}\right ]\cdot \frac {1}{\left [\begin {array}{cccc} -\frac {1}{8} & \frac {1}{8} & 0 & -\frac {1}{8} \\ \frac {1}{4} & \frac {1}{4} & -\frac {1}{4} & 0 \\ -\frac {1}{2} & \frac {1}{2} & 0 & \frac {1}{2} \\ 1 & 1 & 1 & 0 \end {array}\right ]} \\ {} & \circ & \textrm {Evaluate and simplify to get the fundamental matrix}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )=\left [\begin {array}{cccc} \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} & \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} & -\frac {{\mathrm e}^{-2 t}}{32}+\frac {{\mathrm e}^{2 t}}{32}-\frac {\sin \left (2 t \right )}{16} \\ -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} & \frac {{\mathrm e}^{-2 t}}{16}+\frac {{\mathrm e}^{2 t}}{16}-\frac {\cos \left (2 t \right )}{8} \\ {\mathrm e}^{-2 t}+{\mathrm e}^{2 t}-2 \cos \left (2 t \right ) & -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} & -\frac {{\mathrm e}^{-2 t}}{8}+\frac {{\mathrm e}^{2 t}}{8}+\frac {\sin \left (2 t \right )}{4} \\ -2 \,{\mathrm e}^{-2 t}+2 \,{\mathrm e}^{2 t}+4 \sin \left (2 t \right ) & {\mathrm e}^{-2 t}+{\mathrm e}^{2 t}-2 \cos \left (2 t \right ) & -\frac {{\mathrm e}^{-2 t}}{2}+\frac {{\mathrm e}^{2 t}}{2}-\sin \left (2 t \right ) & \frac {{\mathrm e}^{-2 t}}{4}+\frac {{\mathrm e}^{2 t}}{4}+\frac {\cos \left (2 t \right )}{2} \end {array}\right ] \\ \square & {} & \textrm {Find a particular solution of the system of ODEs using variation of parameters}\hspace {3pt} \\ {} & \circ & \textrm {Let the particular solution be the fundamental matrix multiplied by}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {and solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & \circ & \textrm {Take the derivative of the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}^{\prime }\left (t \right )=\Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right ) \\ {} & \circ & \textrm {Substitute particular solution and its derivative into the system of ODEs}\hspace {3pt} \\ {} & {} & \Phi ^{\prime }\left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {The fundamental matrix has columns that are solutions to the homogeneous system so its derivative follows that of the homogeneous system}\hspace {3pt} \\ {} & {} & A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+\Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=A \cdot \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}\left (t \right )+{\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Cancel like terms}\hspace {3pt} \\ {} & {} & \Phi \left (t \right )\cdot {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )={\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Multiply by the inverse of the fundamental matrix}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{v}}^{\prime }\left (t \right )=\frac {1}{\Phi \left (t \right )}\cdot {\moverset {\rightarrow }{f}}\left (t \right ) \\ {} & \circ & \textrm {Integrate to solve for}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right ) \\ {} & {} & {\moverset {\rightarrow }{v}}\left (t \right )=\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \\ {} & \circ & \textrm {Plug}\hspace {3pt} {\moverset {\rightarrow }{v}}\left (t \right )\hspace {3pt}\textrm {into the equation for the particular solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\Phi \left (t \right )\cdot \left (\int _{0}^{t}\frac {1}{\Phi \left (s \right )}\cdot {\moverset {\rightarrow }{f}}\left (s \right )d s \right ) \\ {} & \circ & \textrm {Plug in the fundamental matrix and the forcing function and compute}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{p}\left (t \right )=\left [\begin {array}{c} \frac {\left (\left \{\begin {array}{cc} 1 & t \le 0 \\ -1 & 0

Maple trace

`Methods for high order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 4; linear nonhomogeneous with symmetry [0,1] 
trying high order linear exact nonhomogeneous 
trying differential order: 4; missing the dependent variable 
checking if the LODE has constant coefficients 
<- constant coefficients successful`

\[ y \left (t \right ) = -\frac {\sin \left (2 t \right )}{16}+\frac {\sinh \left (2 t \right )}{16} \]

\[ y(t)\to -\frac {1}{32} e^{-2 t} (\theta (0)-\theta (t)) \left (e^{4 t}-2 e^{2 t} \sin (2 t)-1\right ) \]

22.19 problem 31.7 (L)

22.19.1 Maple step by step solution